Question: Simplify and expand the following expression: $ \dfrac{1}{2k - 2}- \dfrac{3}{5k - 35}- \dfrac{4}{k^2 - 8k + 7} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{1}{2k - 2} = \dfrac{1}{2(k - 1)}$ We can factor a $5$ out of denominator in the second term: $ \dfrac{3}{5k - 35} = \dfrac{3}{5(k - 7)}$ We can factor the quadratic in the third term: $ \dfrac{4}{k^2 - 8k + 7} = \dfrac{4}{(k - 1)(k - 7)}$ Now we have: $ \dfrac{1}{2(k - 1)}- \dfrac{3}{5(k - 7)}- \dfrac{4}{(k - 1)(k - 7)} $ The least common multiple of the denominators is: $ 10(k - 1)(k - 7)$ In order to get the first term over $10(k - 1)(k - 7)$ , multiply by $\dfrac{5(k - 7)}{5(k - 7)}$ $ \dfrac{1}{2(k - 1)} \times \dfrac{5(k - 7)}{5(k - 7)} = \dfrac{5(k - 7)}{10(k - 1)(k - 7)} $ In order to get the second term over $10(k - 1)(k - 7)$ , multiply by $\dfrac{2(k - 1)}{2(k - 1)}$ $ \dfrac{3}{5(k - 7)} \times \dfrac{2(k - 1)}{2(k - 1)} = \dfrac{6(k - 1)}{10(k - 1)(k - 7)} $ In order to get the third term over $10(k - 1)(k - 7)$ , multiply by $\dfrac{10}{10}$ $ \dfrac{4}{(k - 1)(k - 7)} \times \dfrac{10}{10} = \dfrac{40}{10(k - 1)(k - 7)} $ Now we have: $ \dfrac{5(k - 7)}{10(k - 1)(k - 7)} - \dfrac{6(k - 1)}{10(k - 1)(k - 7)} - \dfrac{40}{10(k - 1)(k - 7)} $ $ = \dfrac{ 5(k - 7) - 6(k - 1) - 40} {10(k - 1)(k - 7)} $ Expand: $ = \dfrac{5k - 35 - 6k + 6 - 40}{10k^2 - 80k + 70} $ $ = \dfrac{-k - 69}{10k^2 - 80k + 70}$